[tex]\it Th.\ sinusurilor \Rightarrow \ \dfrac{AC}{sin75^o}=\dfrac{BC}{sin60^o} \Rightarrow AC=BC\cdot\dfrac{sin75^o}{sin60^o}=12\cdot\dfrac{\dfrac{\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt3}{2}}=\\ \\ \\ =12\cdot\dfrac{^{\sqrt3)}\sqrt6+\sqrt2}{\ 2\sqrt3}=12\cdot\dfrac{\sqrt{18}+\sqrt6}{6}=2(3\sqrt2+\sqrt6)[/tex]